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Question: If we have an expression as \[{{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta ,\text{ }{{V}_{n}}=\cos n\t...

If we have an expression as Un=sin⁡nθsec⁡nθ, Vn=cos⁡nθsec⁡nθ≠1{{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta ,\text{ }{{V}_{n}}=\cos n\theta {{\sec }^{n}}\theta \ne 1 , then Vn−Vn−1Un−1+1nUnVn\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}} is equal to
A. 0
B. tan⁡θ\tan \theta
C. −tan⁡θ+tan⁡nθn-\tan \theta +\dfrac{\tan n\theta }{n}
D. tan⁡θ+tan⁡nθn\tan \theta +\dfrac{\tan n\theta }{n}

Explanation

Solution

To find the value of Vn−Vn−1Un−1+1nUnVn\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}} , we will first find Un−1 and Vn−1{{U}_{n-1}}\text{ and }{{V}_{n-1}} by substituting n−1n-1 for nn in Un=sin⁡nθsec⁡nθ and Vn=cos⁡nθsec⁡nθ{{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta \text{ and }{{V}_{n}}=\cos n\theta {{\sec }^{n}}\theta . Then substitute these in Vn−Vn−1Un−1+1nUnVn\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}} ., that is, cos⁡nθsec⁡nθ−cos⁡((n−1)θ)sec⁡n−1θsin⁡((n−1)θ)sec⁡n−1θ+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\cos n\theta {{\sec }^{n}}\theta -\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta } . Solve this using trigonometric identities. We will get sec⁡n−1θ[cos⁡nθsec⁡θ−cos⁡((n−1)θ)]sin⁡((n−1)θ)sec⁡n−1θ+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{{{\sec }^{n-1}}\theta \left[ \cos n\theta \sec \theta -\cos \left( \left( n-1 \right)\theta \right) \right]}{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta } . Solving this further, and writing sec in terms of cos, we will get cos⁡nθ−cos⁡θcos⁡((n−1)θ)cos⁡θsin⁡((n−1)θ)+1nsin⁡nθcos⁡nθ\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta }{\cos n\theta } . Change the equation to use the formula cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) . We will get cos⁡nθ−12[cos⁡nθ+cos⁡((n−2)θ)]cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\cos n\theta -\dfrac{1}{2}\left[ \cos n\theta +\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta . Let us solve this to get 12[cos⁡nθ−cos⁡((n−2)θ)]cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\dfrac{1}{2}\left[ \cos n\theta -\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta . Use the identity cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) and solve to get −tan⁡θ+1ntan⁡nθ-\tan \theta +\dfrac{1}{n}\tan n\theta .

Complete step-by-step solution:
We have to find the value of Vn−Vn−1Un−1+1nUnVn\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}} . It is given that
Un=sin⁡nθsec⁡nθ...(i){{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta ...(i)
Let us substitute n−1n-1 for nn . we will get
Un−1=sin⁡((n−1)θ)sec⁡n−1θ...(ii){{U}_{n-1}}=\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta ...(ii)
It is also given that
Vn=cos⁡nθsec⁡nθ...(iii){{V}_{n}}=\cos n\theta {{\sec }^{n}}\theta ...(iii)
Let us substitute n−1n-1 for nn . we will get
Vn−1=cos⁡((n−1)θ)sec⁡n−1θ...(iv){{V}_{n-1}}=\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta ...(iv)
We have Vn−Vn−1Un−1+1nUnVn\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}}
Now, we can substitute the (i),(ii),(iii) and (iv) in the above equation. That is
cos⁡nθsec⁡nθ−cos⁡((n−1)θ)sec⁡n−1θsin⁡((n−1)θ)sec⁡n−1θ+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\cos n\theta {{\sec }^{n}}\theta -\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
Let us multiply and divide the cos⁡nθsec⁡nθ\cos n\theta {{\sec }^{n}}\theta by sec⁡θ\sec \theta . We will get
cos⁡nθsec⁡nθ×sec⁡θsec⁡θ−cos⁡((n−1)θ)sec⁡n−1θsin⁡((n−1)θ)sec⁡n−1θ+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\cos n\theta {{\sec }^{n}}\theta \times \dfrac{\sec \theta }{\sec \theta }-\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
We know that aman=am−n\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} . Hence, the above equation can be written as
cos⁡nθsec⁡n−1θ×sec⁡θ−cos⁡((n−1)θ)sec⁡n−1θsin⁡((n−1)θ)sec⁡n−1θ+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\cos n\theta {{\sec }^{n-1}}\theta \times \sec \theta -\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
Let us take sec⁡n−1θ{{\sec }^{n-1}}\theta common from the numerator of first term. We will get
sec⁡n−1θ[cos⁡nθsec⁡θ−cos⁡((n−1)θ)]sin⁡((n−1)θ)sec⁡n−1θ+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{{{\sec }^{n-1}}\theta \left[ \cos n\theta \sec \theta -\cos \left( \left( n-1 \right)\theta \right) \right]}{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
Now, we can cancel sec⁡n−1θ{{\sec }^{n-1}}\theta from the numerator and denominator of the first term. That is
cos⁡nθsec⁡θ−cos⁡((n−1)θ)sin⁡((n−1)θ)+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\cos n\theta \sec \theta -\cos \left( \left( n-1 \right)\theta \right)}{\sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
We know that sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } . Hence, the above equation becomes
cos⁡nθcos⁡θ−cos⁡((n−1)θ)sin⁡((n−1)θ)+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\dfrac{\cos n\theta }{\cos \theta }-\cos \left( \left( n-1 \right)\theta \right)}{\sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
This can be written as
cos⁡nθ−cos⁡θcos⁡((n−1)θ)cos⁡θsin⁡((n−1)θ)+1nsin⁡nθsec⁡nθcos⁡nθsec⁡nθ\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }
Meanwhile, let us solve the second term. sec⁡nθ{{\sec }^{n}}\theta can be cancelled from numerator and denominator. We will get
cos⁡nθ−cos⁡θcos⁡((n−1)θ)cos⁡θsin⁡((n−1)θ)+1nsin⁡nθcos⁡nθ\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta }{\cos n\theta }
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } . hence, the above equation becomes
cos⁡nθ−cos⁡θcos⁡((n−1)θ)cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
Let us multiply and divide cos⁡θcos⁡((n−1)θ)\cos \theta \cos \left( \left( n-1 \right)\theta \right) by 2. We will get
cos⁡nθ−22cos⁡θcos⁡((n−1)θ)cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\cos n\theta -\dfrac{2}{2}\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
We know that cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) . Hence, we can write the above form as
cos⁡nθ−12[cos⁡nθ+cos⁡((n−2)θ)]cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\cos n\theta -\dfrac{1}{2}\left[ \cos n\theta +\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
Let us now expand the term 12[cos⁡nθ+cos⁡((n−2)θ)]\dfrac{1}{2}\left[ \cos n\theta +\cos \left( \left( n-2 \right)\theta \right) \right] . We will get
cos⁡nθ−12cos⁡nθ−12cos⁡((n−2)θ)cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\cos n\theta -\dfrac{1}{2}\cos n\theta -\dfrac{1}{2}\cos \left( \left( n-2 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
Solving the numerator of first term, we will get
12cos⁡nθ−12cos⁡((n−2)θ)cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\dfrac{1}{2}\cos n\theta -\dfrac{1}{2}\cos \left( \left( n-2 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
Now let us take 12\dfrac{1}{2} common from the numerator of first term. We will get
12[cos⁡nθ−cos⁡((n−2)θ)]cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\dfrac{1}{2}\left[ \cos n\theta -\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
We know that cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(B−A2)\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)
12[2sin⁡[nθ+(n−2)θ2]sin⁡((n−2)θ−nθ2)]cos⁡θsin⁡((n−1)θ)+1ntan⁡nθ\dfrac{\dfrac{1}{2}\left[ 2\sin \left[ \dfrac{n\theta +\left( n-2 \right)\theta }{2} \right]\sin \left( \dfrac{\left( n-2 \right)\theta -n\theta }{2} \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta
This can be written as

& \dfrac{\dfrac{1}{2}\left[ 2\sin \left( \dfrac{2n\theta -2\theta }{2} \right)\sin \left( \dfrac{-2\theta }{2} \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \\\ & =\dfrac{\dfrac{1}{2}\left[ 2\sin \left( n\theta -\theta \right)\sin \left( -\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \\\ \end{aligned}$$ Let us solve this further by cancelling 2 from the numerator of first term and taking $\theta $ outside from $$\sin \left( n\theta -\theta \right)$$ . $$\dfrac{\sin \left( \left( n-1 \right)\theta \right)\sin \left( -\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta $$ We know that $\sin \left( -\theta \right)=-\sin \theta $ . Hence, the above equation becomes $$\dfrac{\sin \left( \left( n-1 \right)\theta \right)\times -\sin \theta }{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta $$ Let us cancel the common terms from the first term. We will get $$\dfrac{-\sin \theta }{\cos \theta }+\dfrac{1}{n}\tan n\theta $$ We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . Hence, we will get $$-\tan \theta +\dfrac{1}{n}\tan n\theta $$ $$=-\tan \theta +\dfrac{\tan n\theta }{n}$$ Hence, the correct option is C. **Note:** You must know the trigonometric identities and basic conversions to solve this question. You may write the formula for $\cos A+\cos B$ as $\cos A+\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ . This is wrong. You may also make mistake when writing the formula for $\cos A-\cos B$ as $\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ . Also you may write $\sin \left( -\theta \right)=-\cos \theta $ .